What is the derivative of arcsin(1x)?

1 Answer
Antoine
Jul 26, 2015

−1x√x2−1

Explanation:

To differentiate this we will be applying a chain rule:

Start by Letting θ=arcsin(1x)

⇒sin(θ)=1x

Now differentiate each term on both sides of the equation with respect to x

⇒cos(θ)⋅d(θ)dx=−1x2

Using the identity : cos2θ+sin2θ=1⇒cosθ=√1−sin2θ

⇒√1−sin2θ⋅d(θ)dx=−1x2

⇒d(θ)dx=−1x2⋅1√1−sin2θ

Recall : sin(θ)=1x and θ=arcsin(1x)

So we can write,

d(arcsin(1x))dx=−1x2⋅1√1−(1x)2=−1x2⋅1√x2−1x2

=−1x2⋅x√x2−1=−1x√x2−1 or −√x2−1x(x2−1)

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